Calculate the work done on a 1500-kg elevator car by its cable to lift it 40.0 m at constant speed, assuming friction averages 1
00 n. (b) what is the work done on the lift by the gravitational force in this process? (c) what is the total work done on the lift?
1 answer:
The weight of the elevator is
W = (1500 kg)*(9.8 m/s²) = 1.47 x 10⁴ N
Because the frictional resistance is R = 100 N, the tension in the cable for dynamic equilibrium is
T = W + R = 1.47 x 10⁴ + 100 = 1.48 x 10⁴ N
By definition, the work done in lifting the elevator by 40 m is
(1.48 x 10⁴ N)*(40 m) = 5.92 x 10⁵ J = 592 kJ
Answer: 592 kJ (or 5.92 x 10⁵ J)
W = (1500 kg)*(9.8 m/s²) = 1.47 x 10⁴ N
Because the frictional resistance is R = 100 N, the tension in the cable for dynamic equilibrium is
T = W + R = 1.47 x 10⁴ + 100 = 1.48 x 10⁴ N
By definition, the work done in lifting the elevator by 40 m is
(1.48 x 10⁴ N)*(40 m) = 5.92 x 10⁵ J = 592 kJ
Answer: 592 kJ (or 5.92 x 10⁵ J)
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Answer:
1/18
Explanation:
The number of ways in which two die can be thrown is the sample space of the experiment
(1,1), (1,2) (1,3), (1,4) (1,5) (1,6)
(2,1), (2,2) (2,3), (2,4) (2,5) (2,6)
(3,1), (3,2) (3,3), (3,4) (3,5) (3,6)
(4,1), (4,2) (4,3), (4,4) (4,5) (4,6)
(5,1), (5,2) (5,3), (5,4) (5,5) (5,6)
(6,1), (6,2) (6,3), (6,4) (6,5) (6,6)
From here it can be seen that there are only two cases where the sum on the two die is 11 i.e., (6,5) and (5,6)
∴ Probability of getting 11 is<u> 1/18</u>